(c)tan^(-1)|tan x|=|x|

The solution of the differential equation (1+x^2)(dy)/(dx)+1+y^2=0, is a) tan^(-1)x-tan^(-1)y=tan^(-1)C b) tan^(-1)y-tan^(-1)x=tan^(-1)C c) tan^(-1)y+-tan^(-1)x=tan^(\ )C d) tan^(-1)y+tan^(-1)x=tan^(-1)C

The value of int (tan x )/(tan ^(2) x + tan x+1)dx =x -(2)/(sqrtA) tan ^(-1) ((2 tan x+1)/(sqrtA))+C Then the value of A is:

The value of int (tan x )/(tan ^(2) x + tan x+1)dx =x -(2)/(sqrtA) tan ^(-1) ((2 tan x+1)/(sqrtA))+C Then the value of A is:

The value of int (tan x )/(tan ^(2) x + tan x+1)dx =x -(2)/(sqrtA) tan ^(-1) ((2 tan x+1)/(sqrtA))+C Then the value of A is:

The value of 2tan^(-1)(cos ec tan^(-1)x-tan cot^(-1)x) is equal to (a)cot ^(-1)x( b ) (cot^(-1)1)/(x) (c)tan ^(-1)x (d) none of these

int tan^(-1) sqrt x dx=.............a)x tan^(-1) sqrt x - sqrt x + tan^(-1) sqrt x + c b)x tan^(-1) sqrt x + sqrt x - tan^(-1) sqrt x + c c)- x tan^(-1) sqrt x - sqrt x + tan^(-1) sqrt x + c d)x tan^(-1) sqrt x + sqrt x + tan^(-1) sqrt x + c

If tan (x-A)tan (x-B)+tan (x-B)tan (x-C)+tan(x-C)tan (x-A)=1 and A,B and C are the angles of triangle ABC then find x

If tan (x-A)tan (x-B)+tan (x-B)tan (x-C)+tan(x-C)tan (x-A)=1 and A,B and C are the angles of triangle ABC then find x

If int (dx)/(1- sin ^(4)x )= a tan x +b tan ^(-1) (c tan x )+ D, then: